A bar magnet has a pole strength of 15 Am and magnetic length 20 cm. What is the magnetic induction produced by it at a point at a distance of 30 cm from either pole?
`(mu_(0)/(4pi)=10^(-7)" Wb/Am")`

Options

A. `1.11xx10^(-5)" Wb/m"^(2)`
B. `0.8xx10^(-4)" Wb/m"^(2)`
C. `1.5xx10^(-5)" Wb/m"^(2)`
D. `2.75xx10^(-5)" Wb/m"^(2)`

Explanation

To solve the problem, we need to calculate the magnetic induction (B) produced by a bar magnet at a point located at a distance of 30 cm from either pole. We are given the following information:

- Pole strength (m) = 15 Am
- Magnetic length (l) = 20 cm = 0.2 m
- Distance from the pole (d) = 30 cm = 0.3 m
- Value of \(\mu_0/(4\pi) = 10^{-7} \, \text{Wb/Am}\)

### Step-by-step Solution:

1. **Identify the Formula for Magnetic Induction:**
The magnetic induction (B) at a point on the equatorial line of a bar magnet is given by the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{m}{d^2 + l^2}^{3/2} \]

2. **Calculate the Magnetic Moment (M):**
The magnetic moment (M) of the bar magnet can be calculated using the formula:
\[ M = m \cdot 2l \]
Here, \(m = 15 \, \text{Am}\) and \(l = 0.2 \, \text{m}\):
\[ M = 15 \cdot 2 \cdot 0.2 = 6 \, \text{Am}^2 \]

3. **Substitute Values into the Formula:**
Now, we substitute the values into the magnetic induction formula. First, we need to calculate \(d^2 + l^2\):
\[ d = 0.3 \, \text{m}, \quad l = 0.1 \, \text{m} \quad (\text{since } l = 0.2 \, \text{m} \text{ is the total length, half of it is } 0.1 \, \text{m}) \]
\[ d^2 + l^2 = (0.3)^2 + (0.1)^2 = 0.09 + 0.01 = 0.1 \]

4. **Calculate \(d^2 + l^2\) raised to the power of \(3/2\):**
\[ (d^2 + l^2)^{3/2} = (0.1)^{3/2} = 0.1^{1.5} = 0.0316227766 \, \text{m}^{3} \]

5. **Calculate B using the values:**
Now we can substitute all the values into the formula for B:
\[ B = \frac{10^{-7}}{4\pi} \cdot \frac{15}{0.0316227766} \]
\[ B = \frac{10^{-7}}{12.56637061} \cdot 474.341649 \]
\[ B \approx 1.11 \times 10^{-5} \, \text{Wb/m}^2 \]

### Final Answer:
The magnetic induction produced by the bar magnet at a point 30 cm from either pole is approximately:
\[ B \approx 1.11 \times 10^{-5} \, \text{Wb/m}^2 \]